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2 years ago

2 1/3 + 3/7 divided by 2 1/3 x 1 3/7

Mathematics

1 answer:

oee [108]2 years ago

4 0

Answer: $\frac{29}{35}$

Step-by-step explanation:

1. Make sure all your denominators are the same, you can do this by multiplying them all by 3 or 7, 2 and 1/3 would become 2 7/21, 3/7 would become 9/21, 2 and 1/3, would become 2 and 7/21, and so on and so forth, leaving us with

2 $\frac{7}{21}$ + $\frac{9}{21}$ / 2 $\frac{7}{21}$ * 1 $\frac{9}{21}$

2. Now, I would go and solve each side of the equation because we divide them, we can add 2 $\frac{7}{21}$ and $\frac{9}{21}$ together to get 2 $\frac{16}{21}$ , because 7 + 9 = 16

2 $\frac{16}{21}$ / 2 $\frac{7}{21}$ * 1 $\frac{9}{21}$

3. We can do the same thing to the other side, by multiplying the two of them. I'm going to convert both of them to just fractions, 2 $\frac{7}{21}$ becomes $\frac{49}{21}$ and 1 $\frac{9}{21}$ becomes $\frac{30}{21}$

$\frac{49}{21}$ * $\frac{30}{21}$

Now we can reduce them, and for our first fraction divide the top and bottom by 7, giving us $\frac{7}{3}$ and the second one by 3, which gives us $\frac{10}{7}$

$\frac{7}{3}$ * $\frac{10}{7}$ now we cross divide, where we replace the 7's with ones because they're right across from each other to get $\frac{1}{3}$ and $\frac{10}{1}$ , or just 10.

$\frac{1}{3}$ * 10 = $\frac{10}{3}$

4. So now we can do a similar thing with the first half of our equation and convert it into a fraction and not a mixed number.

$\frac{58}{21}$ / $\frac{10}{3}$

Now we multiply divide them, and the easiest way I learned how to do this was by keep the first fraction in its place, and flipping the denominator and numerator of the second, and multiplying them.

so $\frac{58}{21}$ * $\frac{3}{10}$ which if we multiply both sides, you get $\frac{174}{210}$

I'll save you some times and simplify it for you, the greatest common factor is 6 so we divide the top and bottom by 6 to get

$\frac{29}{35}$

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If you're using the app, try seeing this answer through your browser: brainly.com/question/3242555

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Solve the trigonometric equation:

$\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}$

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Let

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So the equation becomes

$\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}$

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$\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.$

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I hope this helps. =)

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